> Observe that there is no prime number which is assigned a nonzero exponent by p_n but not p_m, and there is no prime number which is assigned a nonzero exponent by p_m but not p_n. If p_n assigned a positive exponent to any prime c while p_m assigned c a zero exponent, then the product of p_n would be congruent to 0 (mod c), but the product of p_m would not, and therefore the two products would not equal the same number. (And symmetrically.)Others have taken potshots at this, and I'll join them. Look at the examples (in the article) of Z[sqrt(-5)]:`````` 6 = (2, 0) x (3, 0) => p_n = {(2,0), (3,0)} 6 = (1, 1) x (1, -1) => p_m = {(1,1), (1,-1)} `````` So we have non-uniqueness of the factorization for this case. So we follow the logic of the excerpt above -- c=(2,0) is assigned a zero exponent in the second factorization, and indeed, p_n is congruent to 0 (mod c). Also, p_m is congruent to 0 (mod c). So where do we get the contradiction?The proof of this statement outlined below by makomk does establish this fairly well for the integers, and the argument cannot apply to Z[sqrt(-5)], but only because we can't define an ordering on Z[sqrt(-5)]. Z[i] _is_ a unique factorization domain (despite the lack of ordering), so fundamentally this proof is lacking in its ability to establish unique factorization for a particular ring. Though it does work for the integers, because of the well-ordered property, which is very non-obvious from your proof (and the whole article is about obviousness, not correctness, so this is relevant)

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